00:01
Hi everyone, so what we have is an elevator that's starting from rest and we know that it can accelerate at 5 feet per second squared and decelerate at 2 feet per second squared.
00:10
We know the initial and final velocities are zero and what we're trying to find is the shortest time it takes to get to a position of 40 feet or height of 40 feet.
00:20
So what we can do is we can actually split this into two different sections.
00:24
So the first section will be the acceleration part and then the next section will be the deceleration.
00:30
To the height of 40 feet.
00:32
So what we can say is that v2 will be equal to v1 plus act1.
00:40
So v2 is the velocity at the end of our acceleration portion, and that is t1 is the time it takes to get to finish that acceleration portion.
00:50
So we know our initial velocity is 0.
00:52
We know the elevator accelerates at 5 feet per second squared.
00:56
So we can say that v2 is equal to 5t1.
01:01
Next, v3 will be the velocity of our elevator at the end.
01:09
And that's going to be equal to v2 plus a t2.
01:16
And since we know the elevator is ending at rest, we can see that v3 is equal to 0.
01:22
Therefore, this is going to be equal to v2 minus 2t2.
01:26
Since we know the elevator can decelerate at 2 feet per second square.
01:32
So what this does is that this gives us that t1 is just equal to 0 .4 t2, because we just combine this equation with this equation.
01:46
Now what we can do is go ahead and define our position as well.
01:50
So we can say that s2 is going to be equal to s1 plus v1t1 plus 1 1 -half.
01:58
Act squared and by plugging in the terms that we have and we're using we can say that this is equal to h which will be equal to zero plus zero plus one half five t1 squared so this will just be equal to 2 .5 t1 squared next we can devise a similar equation for the second half of the motion during which the elevator decelerating and say that this is 40 minus h so basically the displacement from our end position to when the acceleration portion ends is equal to 0 plus v max t2 minus 1 half times 2 t2 squared next what we can do is develop an equation for our velocity and say that v squared is equal to v1 squared plus 2ac s minus s1.
03:05
Now, by plugging in the terms that we have, we can say that this is equal to vmax squared is equal to 0 plus 2 times 5, h minus 0.
03:17
So what we can see is that vmax squared is just going to be equal to 10h.
03:23
Similarly, we can also do this for the second part of the motion and say that 0 will be equal to vmax squared since that's the velocity that we're starting, that second part of the motion, the deceleration at, plus 2 minus 2, which is our acceleration, or deceleration, times a displacement, so 40 minus h.
03:44
This we see that vmax squared is equal to 160 minus 4h.
03:52
So as you can see, we have two equations here that give us the value for vmax.
03:57
So we can set these equal to each other and solve for vmax.
04:00
And what we see is that 10h is going to be equal to 160 minus 4h, which gives us the height that the acceleration goes through, or basically the height to which the elevator accelerates to, is equal to 11 .429 feet.
04:20
At this point, the velocity is equal to 10 .69 feet per second...